# Impulse on a Force-Time Graph ## Non-Constant Forces in Real Scenarios - In practice, most forces acting on objects are **not constant** – they change over time - Examples include: - The force exerted by a cricket bat on a ball during impact - The force from a car's brakes during an emergency stop - The thrust from a rocket engine during launch - When forces vary, we need graphical methods to analyse the impulse delivered ## Area Under Force-Time Graphs - When force is plotted against time, **the impulse equals the area under the force-time graph**

Impulse $= \text{area under force-time graph}$

- This applies to any shape of force-time graph: - Constant force → rectangular area - Variable force → curved or angular region - The units work out correctly: - Force (N) × time (s) = N s (units of impulse) ## Why This Works: Mathematical Basis - Starting from the impulse equation: $$\text{Impulse} = F\Delta t$$ - Where: - $F$ = force (N) - $\Delta t$ = time interval (s) - For a constant force, impulse is simply the product $F \times \Delta t$ - For a varying force, we sum many tiny rectangular strips under the graph - This summation gives the total area, which represents the total impulse ## Equivalent Impulse Scenarios - An important consequence: **the same impulse can be delivered by different force-time combinations** - Two equivalent scenarios: - **Small force over long time** → large area from width - **Large force over short time** → large area from height - Example: - A gentle push lasting 10 s might deliver the same impulse as a sharp kick lasting 0.1 s - Both scenarios produce the same change in momentum

Equal areas under force-time graphs mean equal impulses, regardless of the shape

### Example: Constant Force Graph Here's a simple example of a constant force of 50 N applied for 4 seconds. The impulse is the rectangular area = 50 × 4 = 200 N s.

### Example: Variable Force Graph This graph shows a force that increases to a maximum of 80 N at 2 seconds, then decreases back to zero at 4 seconds. The impulse equals the triangular area = ½ × 4 × 80 = 160 N s.

## Types of Force-Time Graphs ### Curved Graphs - Force varies smoothly over time - Common in real collisions where contact builds up gradually then releases - To find the area: - **Count the squares** under the curve - Each square represents a specific impulse value (force unit × time unit) - Partial squares can be estimated (e.g., half-square = 0.5 square) - Total area = (number of squares) × (value per square) ### Straight-Line Graphs - Force changes linearly with time - Graph consists of triangles, rectangles, or trapeziums - To find the area: - **Split the graph into geometric shapes** - Calculate the area of each section using standard formulas - Sum all the individual areas to get total impulse | Shape | Area Formula | Application | |-------|--------------|-------------| | Rectangle | $\text{base} \times \text{height}$ | Constant force region | | Triangle | $\frac{1}{2} \times \text{base} \times \text{height}$ | Linearly increasing/decreasing force | | Trapezium | $\frac{1}{2}(\text{sum of parallel sides}) \times \text{height}$ | Two different constant forces | ## Calculating Areas: Step-by-Step Approach 1. **Identify the time interval** over which the force acts (x-axis extent) 2. **Determine the shape(s)** present in the graph 3. **Apply appropriate formulas** for each geometric section 4. **Sum the areas** if multiple sections exist 5. **Check units** – the result should be in N s ## Worked Example: Rocket Launch A small model rocket of mass 2.5 kg is launched vertically from rest. The thrust force from the engine varies with time as shown in the graph below. The graph consists of two triangular sections: - First triangle: base = 5 s, height = 30 N - Second triangle: base = $(12 - 5) = 7$ s, height = 30 N **Calculate the velocity of the rocket after 12 seconds.**

### Solution **Step 1: Calculate the total impulse** The impulse is the total area under the force-time graph. $$\text{Area of first triangle} = \frac{1}{2} \times 5 \times 30 = 75 \text{ N s}$$ $$\text{Area of second triangle} = \frac{1}{2} \times 7 \times 30 = 105 \text{ N s}$$ $$\text{Total impulse} = 75 + 105 = 180 \text{ N s}$$ **Step 2: Write the impulse-momentum equation** Since the rocket starts from rest, initial momentum = 0 $$\text{Impulse} = \Delta p = mv - mu$$ $$180 = mv - 0$$ $$180 = mv$$ **Step 3: Rearrange to find velocity** $$v = \frac{180}{m} = \frac{180}{2.5}$$ $$v = 72 \text{ m s}^{-1}$$

Always check that your time intervals and force values are read correctly from the graph axes

## Key Geometric Formulas - **Triangle area** = $\frac{1}{2} \times \text{base} \times \text{height}$ - Base = time interval on x-axis - Height = maximum force on y-axis - **Rectangle area** = $\text{width} \times \text{height}$ - Width = time interval - Height = constant force - **Trapezium area** = $\frac{1}{2}(a + b) \times h$ - $a$ and $b$ = parallel sides (forces at start and end) - $h$ = height (time interval) ## Understanding Rate of Change - The concept of "rate of change" is central to impulse calculations - **Rate of change** describes how one quantity varies with respect to another - In physics, rate of change with respect to time is represented by dividing by $\Delta t$ - Examples: - Velocity = rate of change of displacement = $\frac{\Delta s}{\Delta t}$ - Acceleration = rate of change of velocity = $\frac{\Delta v}{\Delta t}$ - Force = rate of change of momentum = $\frac{\Delta p}{\Delta t}$ - The symbol $\Delta t$ represents finite, measurable time intervals - Rearranging $F = \frac{\Delta p}{\Delta t}$ gives $\Delta p = F\Delta t$ (impulse) - Graphically, this product corresponds to the area under the F-t graph

Be comfortable calculating areas of rectangles, triangles, and trapeziums – these appear frequently in force-time graphs

## Practical Tips for Graph Analysis - Always label your axes clearly when sketching graphs in exams - Pay attention to scale factors (e.g., force in kN rather than N) - For complex curves, **estimate partial squares** as fractions (e.g., 0.25, 0.5, 0.75) - When a graph has multiple sections: - Calculate each area separately - Keep track of which section you're working on - Sum all contributions at the end - If the force becomes negative (opposite direction): - The area below the x-axis represents impulse in the opposite direction - Subtract this from positive areas to get net impulse